Talk:2021-04-23 53 9

I am sure you can calculate the probability yourself, but anyway here is my approximation:

Consider a circle of radius r at latitude θ and longitude ϕ. In the latitude/longitude coordinate space, if the circle is small (r ≪ graticule dimensions), it will look like an ellipse with semi-major axis ϕ_r and semi-minor axis θ_r. The semi-major axis will always be longitudinal due to the non-square shapes of the graticules. The lengths of the axes are:

ϕ_r = 1° × r / w_grat,

θ_r = 1° × r / h_grat,

where w_grat and h_grat are the width and the height of the graticule respectively at the centre point of the ellipse. Assuming that the Earth is a sphere with radius R = 6371000 m and taking θ = 53.6401767° from 2021-04-10_53_9, the dimensions of the graticule are approximately

h_grat = π R / 180 = 111195 m,

w_grat = h_grat cos θ = 65922 m.

Now, if r = 384 m, we get the axes of the ellipse:

ϕ_r = 0.00583°,

θ_r = 0.00345°.

The probability P₁ that a random hashpoint hits this ellipse is simply the area of the ellipse divided by the area of the graticule:

P₁ = (π ϕ_r θ_r) / (1° × 1°) = 6.32 × 10⁻⁵ = 1 / 15800.

Finally, the probability P₂ that a hashpoint hits the ellipse at least once in 13 days is

P₂ = 1 − (1 − P₁)¹³ = 8.21 × 10⁻⁴ = 1 / 1220.

An event of this probability will happen on average about once in 3.33 years or three years and four months.

― Kripakko (talk) 13:21, 24 April 2021 (UTC)

Wow, thanks for your calculations! I just used the area of my graticule, which I once calculated as 7299 km² (for spherical Earth), and then worked with ratios, but you actually put in the work with the ellipsis, and I suppose yours is more accurate, though we both disregarded that graticules are actually not rectangular. Anyway, I get (approximately) once in 7299/(0.384²*π*13) days = 3.32 years on average (3.33 with your calculated graticule area).

However, I don't visit every hash, so the probability of it happening for one of the geohashes I've been to this far (and could visit again) is only 1-(1-0.384²*π*13/7299)^76 = 6.1%. That's actually more than I expected! So for equal odds of it happening I'd have to do log(base 1-0.384²*π*13/7299)(0.5) = 840 hashes, which at my current pace would take me about 2.87+(840-78)/37 = 23.5 years total! I don't actually know if my method of calculating these is valid, but it makes sense to me. I just made the formulas up on the spot, since I don't really have any stochastical education.

I love that you're nerdy enough to calculate these as well. I was quite sure you'd notice this as you were the only other geohasher apart from Fippe who went on an expedition on 2021-04-10. Can I put your calculations on the expedition page? I think they'll be more appreciated there. --Π π π (talk) 10:10, 25 April 2021 (UTC)